Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.util.*;
public class a{
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int N = in.nextInt();
		int M = in.nextInt();
		int[][] map = new int[N][M];
		int min = 2147483647;
		for(int n=0;n<N;n++){
			for(int m=0;m<M;m++){
				map[n][m] = in.nextInt();
				min = Math.min(min, map[n][m]);
			}
		}
		
		// make it so that there is at least one zero on the board
		for(int n=0;n<N;n++){
			for(int m=0;m<M;m++){
				map[n][m] -= min;
			}
		}
		
		// search for this zero now
		int nZero = -1, mZero = -1;
		for(int n=0;n<N;n++){
			for(int m=0;m<M;m++){
				if(map[n][m] == 0){
					nZero = n;
					mZero = m;
					break;
				}
			}
			if(nZero != -1) break;
		}
		
		// that's the position of the first zero, okay
		// but what about the other columns?
		
		// stands for "homogeneous columns"
		int[][] homoColumns = new int[N][M];
		for(int n=0;n<N;n++) homoColumns[n] = map[n].clone();
		for(int n=0;n<N;n++){
			for(int m=0;m<M;m++){
				if(mZero == m) continue;
				homoColumns[n][m] -= map[n][mZero];
			}
			homoColumns[n][mZero] = 0;
		}
		
		// stands for "homogeneous rows"
		int[][] homoRows = new int[N][M];
		for(int n=0;n<N;n++) homoRows[n] = map[n].clone();
		for(int m=0;m<M;m++){
			for(int n=0;n<N;n++){
				if(nZero == n) continue;
				homoRows[n][m] -= map[nZero][m];
			}
			homoRows[nZero][m] = 0;
		}
		
//		for(int n=0;n<N;n++) System.out.println(Arrays.toString(homoRows[n]));
//		System.out.println();
//		for(int[] a : homoColumns) System.out.println(Arrays.toString(a));
//		
		// are all the rows/cols the same?
		boolean fail = false;
		for(int n=0;n<N;n++){
			for(int m=1;m<M;m++){
				if(homoRows[n][m-1] != homoRows[n][m]){
					fail = true;
				}
			}
		}
		for(int n=1;n<N;n++){
			for(int m=0;m<M;m++){
				if(homoColumns[n-1][m] != homoColumns[n][m]){
					fail = true;
				}
			}
		}
		if(fail){
			System.out.println(-1);
			return;
		}
		else{
			int times = Math.min(N, M) * min;
			for(int n=0;n<N;n++){
				times += homoRows[n][0];
			}
			for(int m=0;m<M;m++){
				times += homoColumns[0][m];
			}
			
			System.out.println(times);
			// min has to be taken into account
			for(; min > 0; min--){
				if(N > M){
					for(int m=0;m<M;m++){
						System.out.println("col "+(m+1));
					}
				}
				else{
					for(int n=0;n<N;n++){
						System.out.println("row "+(n+1));
					}
				}
			}
			
			// now the rest of it
			for(int n=0;n<N;n++){
				for(; homoRows[n][0] > 0; homoRows[n][0]--){
					System.out.println("row "+(n+1));
				}
			}
			for(int m=0;m<M;m++){
				for(; homoColumns[0][m] > 0; homoColumns[0][m]--){
					System.out.println("col "+(m+1));
				}
			}
		}
	}
}